Calculate the volume of oxygen gas, measured at 30.0°C and 725 torr that can be produced from complete decomposition of 3.49 grams of mercury(II) oxide.

Answer 1

2HgO ==> 2Hg + O₂(g) .. balanced equation for decomposition of mercury(II) oxide

molar mass HgO = 216.6 g / mol

mols HgO present = 3.49 g HgO x 1 mol HgO / 216.6 g = 0.01611 mols HgO

mols O2 produced = 0.01611 mols HgO x 1 mol O₂ / 2 mol HgO = 0.008056 mols O₂

To convert this to volume, use the ideal gas law PV = nRT

PV = nRT

V = nRT / P

V = volume in liters = ?

n = mols O₂ = 0.008056

R = gas constant = 62.36 Ltorr/Kmol

T = temperature in Kelvin = 30C + 273 = 303K

P = pressure = 725 torr

V = (0.008056)(62.36)(303) / 725

V = 0.210 L